====== Question 5(i) & 5(ii) Exercise 3.5 ======
Solutions of Question 5(i) & 5(ii) of Exercise 3.5 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

=====Question 5(i)=====
Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\quad$ and $\quad\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\quad$. Find $\vec{a} \times \vec{b}\quad$ and prove that $\vec{a} \times \vec{b}$ is orthogonal to both $\vec{a}$ and $\vec{b}$
====Solution====
To show that $\vec{a} \times \vec{b}$ is orthogonal to both $\vec{a}$ and $\vec{b}$.

We check the dot product of $\vec{a} \times \vec{b}$ with $\vec{a}$ and $\vec{b}$.

For $\vec{a} \times \vec{b}$ orthogonal to $\vec{a}$

\begin{align}\vec{a} \cdot \vec{a} \times \vec{b}&=\left|\begin{array}{lll}
a_1 & a_2 & a_3 \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 
\end{array}\right|\\
&=0\quad \because \text{two rows are identical}\\
\Rightarrow \quad \vec{a} \cdot \vec{a} \times \vec{b}&=0\end{align}\\
Which implies that $\vec{a} \times \vec{b} \perp \vec{a}$.

For $\vec{a} \times \vec{b}$ orthogonal to $\vec{b}$

\begin{align}\vec{b} \cdot \vec{a} \times \vec{b}&=\left|\begin{array}{lll}
b_1 & b_2 & b_3 \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{array}\right|\\
&=0\quad\because \quad\text{ two rows are identical}\\
\Rightarrow \vec{a} \cdot(\vec{a} \times \vec{b})&=0.\end{align}
Which implies that $\vec{a} \times \vec{b}\perp \vec{b}$.

=====Question 5(ii)=====
Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$ and $\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$. Find $\vec{a} \times \vec{b}$ and  $|\vec{a} \times \vec{b}|^2$
====Solution====
We know that
\begin{align}\vec{a}\times \vec{b}&=\left|\begin{array}{lll}
\hat{i}&\hat{j}&\hat{k}\\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 
\end{array}\right|\\
\vec{a} \times \vec{b}&=(a_2 b_3-a_3 b_2) \hat{i} -(a_1 b_3-a_3 b_1) \hat{j}+(a_1 b_2-a_2 b_1)\hat{k} \\
\Rightarrow|\vec{a} \times \vec{b}|&=\sqrt{ (a_2 b_3-a_3 b_2)^2+(a_1b_3- a_3 b_1)^2+(a_1 b_2-a_2 b_1)^2} \end{align}
Taking square of the both sides,
$$|\vec{a} \times \vec{b}|^2=(a_2 b_3 - a_3 b_2)^2+(a_1 b_3-a_3 b_1)^2 +(a_1 b_2-a_2 b_1)^2.$$




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